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STRUCTURE OF Sc-45 AND Ti-45
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity. In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications . Structure of stable Sc-45 with S = -7/2 The stable Sc-45 consists of 21p and 24n. Thus it has one additional proton and four additional neutrons than Ca-40. ( See my STRUCTURE OF Ca-40 AND Ca-48 ). For example Ca-40 (from p1 to n20) has 20 protons and 20 neutrons with S=0, while the 21 protons and the 24 neutrons of Sc-45 give the total spin S= -7/2. In the following diagram one observes the additional deuteron of S = -1 having the p21(-1/2) and n21(-1/2). Then the p15(+1/2) which forms the structure of alpha particle in Ca-40 is moved from the third horizontal plane to the second horizontal plane with spin S = -1/2. Note that this transformation from S= +1/2 to S =-1/2 gives a spin S = -1. In this new position it makes not only the two (p15-n4) and (p15-n15) bonds but also the (p15-n22) bond, because there exists the new extra neutron n22(-1/2). Also the p21(-1/2) makes the three pn bonds like (p21-n3), (p21-n13), and (p21-n21) which can overcome the pp repulsions of long range. So the Cs-45 is a stable nuclide. On the other hand the deuteron of S=-1, the transformation of S=-1, and the n22(-1/2) give S = -5/2. Therefore adding the neutrons n23(-1/2) and n24(-1/2) one gets the structure of Sc-45 with S=-7/2. Such extra neutrons fill the blank positions between the protons for making two np bonds per neutron. It is of interest to note that the blank positions between protons formed by the α particles in front of the Mg-24 or behind it, make blank positions able to receive extra protons. However the two pn bonds per proton cannot overcome the pp repulsions of long range. Fortunately in this case the extra n23(-1/2) and n24(-1/2) are able to overcome the nn repulsions of short range. That is, the extra two np bonds per neutron lead to the stability of Sc-45. They belong to the second horizontal plane but on purpose they are not shown here. Note that the nucleons which belong to the second horizontal plane have negative spins (-HP2). To conclude comparing the structure of Sc-45 with the structure of Ca-40 we see that the total spin S = -7/2 of Sc-45 is due to the transformation from p15(+1/2) of Ca-40 to p15(-1/2) of Sc-45 which gives S =-1 and to the five additional nucleons as p21, n21, n22, n23, and n24 which belong to the second horizontal plane giving S = -5/2. Thus the total spin of Sc-45 is S = 0 -1 -5/2 = -7/2 ' ' ' ' Structure of unstable Ti-45 with S = -7/2 The unstable Ti-45 consists of 22 protons and 23 neutrons. It means that the n24(-1/2) of the blank position between two protons is replaced by the new p22(-1/2) filling the neighboring blank position between two neutrons. Consequently Ti-45 has the same spin S =-7/2 as that of Sc-45. However since p22 makes two pn bonds per proton it is unable to overcome the repulsive energy Q of pp repulsions of long range. Under this condition the p22(-1/2) decays into the n24(-1/2) for making the stable Sc-45 with the same spin S = -7/2. ' ' ' STRUCTURE OF STABLE Sc-45 WITH S = -7/2' (The structure consists of 6 horizontal planes existing from +HP1 with positive spins to the -HP6 of negative spins. The nucleons p17, n17, p18, n18, p19, n19, p20 and n20 of the two 'α'' particles in front of Mg-24 and behind it along with the extra neutrons as n23 and n24 are not shown here)' ' ' ' p12..........n12' ' n11..........p11 -HP6' ' n10..........p10' ' p9............n9 +HP5' ' n14..........p8............n8............p16' ' p14..........n7............p7...........n16 -HP4' ' p13..........n6............p6............n15' ' n13..........p5...........n5 +HP3' ' n21.........p4............n4............p15' ' p21..........n3............p3............n22 -HP2' ' n2............p2' ' p1...........n1 +HP1' Category:Fundamental physics concepts